How can I find the number of arguments of a Python function?

Question

How can I find the number of arguments of a Python function? I need to know how many normal arguments it has and how many named arguments.

Example:

def someMethod(self, arg1, kwarg1=None):
    pass

This method has 2 arguments and 1 named argument.

Answer 1

The previously accepted answer has been deprecated as of Python 3.0. Instead of using inspect.getargspec you should now opt for the Signature class which superseded it.

Creating a Signature for the function is easy via the signature function:

from inspect import signature

def someMethod(self, arg1, kwarg1=None):
    pass

sig = signature(someMethod)

Now, you can either view its parameters quickly by string it:

str(sig)  # returns: '(self, arg1, kwarg1=None)'

or you can also get a mapping of attribute names to parameter objects via sig.parameters.

params = sig.parameters 
print(params['kwarg1']) # prints: kwarg1=20

Additionally, you can call len on sig.parameters to also see the number of arguments this function requires:

print(len(params))  # 3

Each entry in the params mapping is actually a Parameter object that has further attributes making your life easier. For example, grabbing a parameter and viewing its default value is now easily performed with:

kwarg1 = params['kwarg1']
kwarg1.default # returns: None

similarly for the rest of the objects contained in parameters.

---

As for Python 2.x users, while inspect.getargspec isn't deprecated, the language will soon be :-). The Signature class isn't available in the 2.x series and won't be. So you still need to work with inspect.getargspec.

As for transitioning between Python 2 and 3, if you have code that relies on the interface of getargspec in Python 2 and switching to signature in 3 is too difficult, you do have the valuable option of using inspect.getfullargspec. It offers a similar interface to getargspec (a single callable argument) in order to grab the arguments of a function while also handling some additional cases that getargspec doesn't:

from inspect import getfullargspec

def someMethod(self, arg1, kwarg1=None):
    pass

args = getfullargspec(someMethod)

As with getargspec, getfullargspec returns a NamedTuple which contains the arguments.

print(args)
FullArgSpec(args=['self', 'arg1', 'kwarg1'], varargs=None, varkw=None, defaults=(None,), kwonlyargs=[], kwonlydefaults=None, annotations={})

Answer 2

import inspect
inspect.getargspec(someMethod)

see the inspect module

Answer 3

someMethod.func_code.co_argcount

or, if the current function name is undetermined:

import sys

sys._getframe().func_code.co_argcount

Answer 4

inspect.getargspec()

Get the names and default values of a function’s arguments. A tuple of four things is returned: (args, varargs, varkw, defaults). args is a list of the argument names (it may contain nested lists). varargs and varkw are the names of the * and ** arguments or None. defaults is a tuple of default argument values or None if there are no default arguments; if this tuple has n elements, they correspond to the last n elements listed in args.

Changed in version 2.6: Returns a named tuple ArgSpec(args, varargs, keywords, defaults).

See can-you-list-the-keyword-arguments-a-python-function-receives.

Answer 5

func.__code__.co_argcount gives you number of any arguments BEFORE *args

func.__kwdefaults__ gives you a dict of the keyword arguments AFTER *args

func.__code__.co_kwonlyargcount is equal to len(func.__kwdefaults__)

func.__defaults__ gives you the values of optional arguments that appear before *args

Here is the simple illustration:

>>> def a(b, c, d, e, f=1, g=3, h=None, *i, j=2, k=3, **L):
    pass

>>> a.__code__.co_argcount
7
>>> a.__defaults__
(1, 3, None)
>>> len(a.__defaults__)
3
>>> 
>>> 
>>> a.__kwdefaults__
{'j': 2, 'k': 3}
>>> len(a.__kwdefaults__)
2
>>> a.__code__.co_kwonlyargcount
2

Answer 6

Adding to the above, I've also seen that the most of the times help() function really helps

For eg, it gives all the details about the arguments it takes.

help(<method>)

gives the below

method(self, **kwargs) method of apiclient.discovery.Resource instance
Retrieves a report which is a collection of properties / statistics for a specific customer.

Args:
  date: string, Represents the date in yyyy-mm-dd format for which the data is to be fetched. (required)
  pageToken: string, Token to specify next page.
  parameters: string, Represents the application name, parameter name pairs to fetch in csv as app_name1:param_name1, app_name2:param_name2.

Returns:
  An object of the form:

    { # JSON template for a collection of usage reports.
    "nextPageToken": "A String", # Token for retrieving the next page
    "kind": "admin#reports#usageReports", # Th

Answer 7

Good news for folks who want to do this in a portable way between Python 2 and Python 3.6+: use inspect.getfullargspec() method. It works in both Python 2.x and 3.6+

As Jim Fasarakis Hilliard and others have pointed out, it used to be like this:
1. In Python 2.x: use inspect.getargspec()
2. In Python 3.x: use signature, as getargspec() and getfullargspec() were deprecated.

However, starting Python 3.6 (by popular demand?), things have changed towards better:

From the Python 3 documentation page:

inspect.getfullargspec(func)

Changed in version 3.6: This method was previously documented as deprecated in favour of signature() in Python 3.5, but that decision has been reversed in order to restore a clearly supported standard interface for single-source Python 2/3 code migrating away from the legacy getargspec() API.

Answer 8

inspect.getargspec() to meet your needs

from inspect import getargspec

def func(a, b):
    pass
print len(getargspec(func).args)

Answer 9

As other answers suggest, getargspec works well as long as the thing being queried is actually a function. It does not work for built-in functions such as open, len, etc, and will throw an exception in such cases:

TypeError: <built-in function open> is not a Python function

The below function (inspired by this answer) demonstrates a workaround. It returns the number of args expected by f:

from inspect import isfunction, getargspec
def num_args(f):
  if isfunction(f):
    return len(getargspec(f).args)
  else:
    spec = f.__doc__.split('\n')[0]
    args = spec[spec.find('(')+1:spec.find(')')]
    return args.count(',')+1 if args else 0

The idea is to parse the function spec out of the __doc__ string. Obviously this relies on the format of said string so is hardly robust!

Answer 10

In:

import inspect 

class X:
    def xyz(self, a, b, c): 
        return 

print(len(inspect.getfullargspec(X.xyz).args))

Out:

4

---

Note: If xyz wasn't inside class X and had no "self" and just "a, b, c", then it would have printed 3.

For python below 3.5, you may want to replace inspect.getfullargspec by inspect.getargspec in the code above.

Answer 11

The accepted answer by Dimitris Fasarakis Hilliard suggests getting parameters in the string format but I think one can make a mistake when parsing this string and thus I created rather a list of the parameters directly using the inspect module

import inspect
def my_function(a,b,c):
    #some code
    pass

result=list(inspect.signature(my_function).parameters.keys())
print(result)
['a','b','c']

Answer 12

Assuming you may be dealing with class based methods or simply functions, you could do something like the following.

This will automatically subtract one input if the input is a class method (and therefore includes self).

import types
def get_arg_count(fn):
    extra_method_input_count=1 if isinstance(fn, types.MethodType) else 0
    return fn.__code__.co_argcount-extra_method_input_count

Then you can apply as you need to functions or methods:

def fn1(a, b, c):
    return None

class cl1:
    def fn2(self, a, b, c):
        return None

print(get_arg_count(fn1)) #=> 3
print(get_arg_count(cl1().fn2)) #=> 3